I’m not sure about:
1) Which voltage 12 or 24.
2) How many amps
3) How many watts
Basically all it does is, cools water (from 50 deg C) so it can be drinkable (filtering has been sorted out). Again, the tank ~ holds 65 liters.
I’m not sure about:
1) Which voltage 12 or 24.
2) How many amps
3) How many watts
Basically all it does is, cools water (from 50 deg C) so it can be drinkable (filtering has been sorted out). Again, the tank ~ holds 65 liters.
Presumably the panel is used to power a pump. The pump can be specified if you know the suction head, discharge head and flow. Simplistically, add the two heads, add some extra like 30% for pipe resistance to the head. This assumes proper pipe-size and a short distance. Now use the formula:
Power_W = (Mass_kg x Gravity_m/s/s x Height_m) / time_s
The earth’s gravity is 9.8m/s/s.
One liter is 1Kg, so work from the desired flow in liters per second, and the total head in meters.
As you can see the power is about how much water is lifted, how far and how quickly.
Now add 30% for pump and motor losses, and that is the power for the pump. If the power is more than about 100W think of a 24V system. A 12V panel will suit otherwise. If you want the pump to just run when there is sun, the panel needs to be this power or more. The pump draws only what it needs. You may need a suitable voltage regulator for the pump, and some sort of protection for the panel. Look into solar powered pumps.
This will only work when the full sun is on the panel. It may make sense to have a battery and charger as well. In this case add an extra 40% for battery charging losses (multiply by 1.4). A lower power panel could be sufficient if the pump only runs some of the time. If it runs at times without the sun, or 24 x 7 you must have a battery, and the panel will need to be bigger.
Calculate the energy in Wh per day from
hours_operating_per_day x pump_and_battery_charging_power.
Divide this by the number of full sun equivalent hours per day in your region, maybe 2 to 5h a day. This gives the panel power. Determine current from the power divided by from the battery voltage. The size of the battery comes from the current and the time it needs to run, in ampere hours. Allow extra for losses.
The panels are usually designated as 12 or 24V models, designed to charge 12 or 24V batteries. The power is the maximum power they can provide. Look on the data sheet, showing the voltage and the current at maximum power. The power is also affected by temperature, as the voltage is reduced as panels heat up considerably in the sun. The rating is only with the panel square on facing the full sun. Any other angle of x or y reduces power by cosine(theta) so at 90 degrees = zero.
The best way to attack this is to figure out what flow rate of water you need, then select a pump based on that. A 1 liter/minute pump will fill the tank in about an hour. If you only need to fill the tank once a day, you might get away with even less than that. That would be just a tiny fountain pump unless you had to raise the water to significant height. You would want a food grade pump and tubing if this is for drinking water.
The people here http://www.wind-sun.com/ForumVB/forumdisplay.php?f=10 know a lot about solar water pumping.